What is the largest integer n&lt;0 so that (1+33i)n is purely imaginary?

n=−1 This is the largest negative integer, and (1+33i)−1=3i1+3 is clearly purely imaginary.

Well the largest integer n&lt;0 is -1 and as it happens that gives an imaginary result.(1+33i)−1=3i1+3 and as 11+3 is a Real Number it follows that this 3i is purely imaginary.If you were actually thinking that -2 is a ''larger'' negative number, and -3 larger still then the question doesn't really make sense because any odd (negative) power of (1+33i) is also purely imaginary, so there would be no ''largest'' negative n in that sense.

n=−1 This is the largest negative integer, and (1+33i)−1=3i1+3 is clearly purely imaginary.

Well the largest integer n&lt;0 is -1 and as it happens that gives an imaginary result.(1+33i)−1=3i1+3 and as 11+3 is a Real Number it follows that this 3i is purely imaginary.If you were actually thinking that -2 is a ''larger'' negative number, and -3 larger still then the question doesn't really make sense because any odd (negative) power of (1+33i) is also purely imaginary, so there would be no ''largest'' negative n in that sense.

Find out the answer to "What is the largest integer n<0 so that..."

Bobbie Comstock

Answered question

2022-01-18

What is the largest integer

n < 0

so that

{(\frac{1 + \sqrt{3}}{3} i)}^{n}

is purely imaginary?

Answer & Explanation

ol3i4c5s4hr

Beginner2022-01-19Added 48 answers

n = - 1

This is the largest negative integer, and

{(\frac{1 + \sqrt{3}}{3 i})}^{- 1} = \frac{3 i}{1 + \sqrt{3}}

is clearly purely imaginary.

censoratojk

Beginner2022-01-20Added 46 answers

Well the largest integer

n < 0

is -1 and as it happens that gives an imaginary result.

{(\frac{1 + \sqrt{3}}{3} i)}^{- 1} = 3 \frac{i}{1 + \sqrt{3}}

and as

\frac{1}{1 + \sqrt{3}}

is a Real Number it follows that this 3i is purely imaginary.
If you were actually thinking that -2 is a ''larger'' negative number, and -3 larger still then the question doesn't really make sense because any odd (negative) power of

(\frac{1 + \sqrt{3}}{3} i)

is also purely imaginary, so there would be no ''largest'' negative n in that sense.

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